Jak obliczyć sin^2 10^o + sin^2 80^o ? prosze o pomoc, pilne :)

Zauważ, że   [latex]sin(90^{circ}-alpha) = cos alpha quad hbox{stad:} \ sin80^{circ}=sin(90^{circ}-10^{circ})=cos10^{circ}[/latex]. Czyli: [latex]sin^{2}10^{circ}+sin^{2}80^{circ}=sin^{2}10^{circ}+cos^{2}10^{circ}=1 \ hbox{bo} quad sin^{2}x+cos^{2}x=1[/latex].